OPS435 Python Assignment 2 A

Overview

When making back up of data files or log files, it is a very common practice to name the backup directories and/or files based on the date the backup was done. In order to restore or locate the directory/file, we often need to find out the backup date from today's date.

The task for this assignment is to write a python program which will take a date in the "YYYYMMDD" format and the number of day before or after the given date as the command line arguments, and output the actual date which is the number of day before or after the given date in the same format.

Instruction

Program Name and valid command line arguments

Name your python3 program as dbda.py. The program should accept two command line parameters, the first one is the date in "YYYYMMDD" format, and the second one is the number of day from the given date, a position value indicates the number of days after the given date, and a negative value indicates the number of days before the given date. For examples:

• python3 dbda.py 20180101 1, and the output should be
  

    20180102

• python3 dbda.py 20180101 -1, and the output should be
  

    20171231

• python3 dbda.py 20180101 2, and the output should be
  

    20180103

If there is too few or too many command line parameters given, display the proper usage.

Program structure

Your program code should all be in a single python file with at least two functions, one called tomorrow() and the other called yesterday().

Sample code for the tomorrow() function

# echo the date in YYYYMMDD after the current day
# 
def tomorrow(today):
    if len(today) != 8:
       return '00000000'
    else:
       year = int(today[0:4])
       month = int(today[4:6])
       day = int(today[6:])

       lyear = year % 4
       if lyear == 0:
          feb_max = 29 # this is a leap year
       else:
          feb_max = 28 # this is not a leap year

       lyear = year % 100
       if lyear == 0:
          feb_max = 28 # this is not a leap year

       lyear = year % 400
       if lyear == 0:
          feb_max = 29 # this is a leap year

       tmp_day = day + 1 # tomorrow's day

       mon_max = { 1:31, 2:feb_max, 3:31, 4:30, 5:31, 6:30, 7:31, 8:31, 9:30, 10:31, 11:30, 12:31}
       if tmp_day > mon_max[month]:
          to_day = tmp_day % mon_max[month] # if tmp_day > this month's max, reset to 1
          tmp_month = month + 1
       else:
          to_day = tmp_day
          tmp_month = month + 0

       if tmp_month > 12:
           to_month = 1
           year = year + 1
       else:
           to_month = tmp_month + 0

       tomorrow_date = str(year)+str(to_month).zfill(2)+str(to_day).zfill(2)
     
       return tomorrow_date